Let $x$ represent the proportion of employees in the company who are in their first five years, and $1 - x$ represent the proportion of employees who have been there for more than five years.

The information provided can be translated into the following:

• 10% of employees in their first five years earn 100,000 or more. This means $0.10x$ represents the proportion of the total workforce who are in their first five years and earn 100,000 or more.
• 20% of employees who have been with the company for more than five years earn 100,000 or more. This is $0.20(1 - x)$ representing the proportion of the total workforce in this category earning 100,000 or more.
• Overall, 13% of the entire workforce earns 100,000 or more, which is given as $0.13$.

The total percentage of employees who earn 100,000 or more can be expressed as the sum of the two groups:

$0.10x + 0.20(1 - x) = 0.13$

Now, let's solve for $x$:

$0.10x + 0.20 - 0.20x = 0.13$

Simplify:

$-0.10x + 0.20 = 0.13$

Subtract 0.20 from both sides:

$-0.10x = 0.13 - 0.20$ $-0.10x = -0.07$

Divide by -0.10:

$x = \frac{-0.07}{-0.10} = 0.7$

Thus, the proportion of employees in their first five years is $\boxed{0.7}$ or 70%.

Would you like further details or clarification on any part of the solution?

Here are some additional related questions:

1. How would the result change if 15% of new employees earned 100,000 or more?
2. What is the probability that a random employee earning 100,000 or more is in their first five years?
3. How would the company’s total percentage of high earners change if the number of employees in their first five years increased by 5%?
4. What if the percentage of employees earning 100,000 or more after their first five years increased to 25%?
5. How could this analysis be adapted to track employees earning 150,000 or more?

Tip: Always set up a clear relationship between given percentages and unknown quantities using algebraic variables for efficient problem-solving.